i.e \(probability = \frac{5C3}{10C3} + \frac{5C3}{10C3}\) (we have taken two times 5C3 because we need to have exclusive to one gender i.e either men or women, so combining them will give us total number of favorable events)

\(= \frac{1}{12} +\frac{1}{12} =\frac{1}{6}\)
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